Section 12.8 : Tangent, Normal and Binormal Vectors
1. Find the unit tangent vector for the vector function : \(\vec r\left( t \right) = \left\langle {{t^2} + 1,3 - t,{t^3}} \right\rangle \)
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Start SolutionFrom the notes in this section we know that to get the unit tangent vector all we need is the derivative of the vector function and its magnitude. Here are those quantities.
\[\vec r'\left( t \right) = \left\langle {2t, - 1,3{t^2}} \right\rangle \] \[\left\| {\vec r'\left( t \right)} \right\| = \sqrt {{{\left( {2t} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( {3{t^2}} \right)}^2}} = \sqrt {1 + 4{t^2} + 9{t^4}} \] Show Step 2The unit tangent vector for this vector function is then,
\[\vec T\left( t \right) = \frac{1}{{\sqrt {1 + 4{t^2} + 9{t^4}} }}\left\langle {2t, - 1,3{t^2}} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle {\frac{{2t}}{{\sqrt {1 + 4{t^2} + 9{t^4}} }}, - \frac{1}{{\sqrt {1 + 4{t^2} + 9{t^4}} }},\frac{{3{t^2}}}{{\sqrt {1 + 4{t^2} + 9{t^4}} }}} \right\rangle }}\]